Oxidation state of sulphur in tetrathionic ac | vedclass

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Question

(D) The chemical formula for tetrathionic acid is $H_2S_4O_6$. Its structure is $HO-SO_2-S-S-SO_2-OH$. In this structure,the two central sulfur atoms

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$+5$ and $0$

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(D) The chemical formula for tetrathionic acid is $H_2S_4O_6$. Its structure is $HO-SO_2-S-S-SO_2-OH$. In this structure,the two central sulfur atoms are bonded only to other sulfur atoms,so their oxidation state is $0$. The two terminal sulfur atoms are each bonded to three oxygen atoms and one sulfur atom. Calculating the oxidation state for terminal sulfur $(x)$: $x + 3(-2) + 1(-1) = -1$ (considering the $S-OH$ bond),or more simply,by formal charge assignment,the terminal sulfur atoms have an oxidation state of $+5$. Thus,the oxidation states of sulfur in $H_2S_4O_6$ are $+5$ and $0$.

Oxidation state of sulphur in tetrathionic acid is

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